(a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares

Exercises · 9.23

Q. (a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power?

(b) What is the magnification in this case?

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Concept understanding — Microscope Magnification

Microscope Magnification

A compound microscope views very small, nearby objects using two lenses in sequence: the objective (near the object) and the eyepiece (near the eye). Its total magnifying power is the product of what each lens contributes.

How the two lenses work together

The object sits just beyond the focus of the short-focal-length objective ( $f_o$ ), which forms a real, inverted, magnified image inside the tube.
That real image falls just inside the focus of the eyepiece ( $f_e$ ), which acts as a simple magnifier, producing a large virtual, magnified final image for the eye.

Because each stage magnifies, the effects multiply:

M = m_o \times m_e

The objective's magnification

m_o = \frac{v_o}{u_o} \approx \frac{L}{f_o}

where $L$ is the tube length (roughly the distance between the objective's second focal point and the eyepiece's first focal point). The object sits close to $f_o$ , so this approximation holds for a well-designed microscope.

The eyepiece's magnification

The eyepiece behaves as a simple magnifier:

Final image at the near point ( $D = 25\ \text{cm}$ , largest magnification):

m_e = 1 + \frac{D}{f_e}

Final image at infinity (relaxed eye, "normal adjustment"):

m_e = \frac{D}{f_e}

Total magnifying power

M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)

Important

High magnification needs short $f_o$ and $f_e$ (both sit in denominators) and a large tube length $L$ — this is why a microscope objective is always a very short-focus lens.

Worked example

Objective $f_o = 1.0\ \text{cm}$ , eyepiece $f_e = 2.5\ \text{cm}$ , tube length $L = 20\ \text{cm}$ , near point $D = 25\ \text{cm}$ . Find $M$ with the final image at the near point.

m_o = \frac{20}{1.0} = 20, \qquad m_e = 1 + \frac{25}{2.5} = 11

$M = 20 \times 11 = 220$ …

Why this formula?

Microscope Magnification: Why the Formula Holds

Let's build the understanding from first principles — not just memorizing formulas, but seeing why they work.

1. What Does "Magnification" Mean in a Microscope?

A microscope creates a larger apparent image of a tiny object. The total magnification is the product of two stages:

Objective lens — creates a real, enlarged, inverted image of the specimen.
Eyepiece (ocular) — acts as a simple magnifier to view that real image.

So:

Total magnification = (magnification by objective) × (magnification by eyepiece)

2. The Key Formula

For a compound microscope in normal adjustment (final image at infinity, relaxed eye):

M = M_o \times M_e = \left( \frac{L}{f_o} \right) \times \left( \frac{D}{f_e} \right)

Where:

$f_o$ = focal length of objective
$f_e$ = focal length of eyepiece
$L$ = tube length (distance between second focal point of objective and first focal point of eyepiece)
$D$ = near point distance of the eye (usually 25 cm)

3. Derivation of Objective Magnification $M_o$

Step 1: How the objective works

The objective lens forms a real, inverted, enlarged image of the specimen. The specimen is placed just outside its focal point ( $f_o$ ).

Step 2: Using the lens formula

For a thin lens:

\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}

(Using sign convention: $u_o$ is negative, $v_o$ is positive)

Step 3: The tube length approximation

In a standard microscope, the specimen is placed very close to $f_o$ , so:

$u_o \approx -f_o$ (object just beyond focal point)
The image is formed at the first focal point of the eyepiece, which is at a distance $L$ from the second focal point of the objective.

Thus:

v_o \approx f_o + L

Step 4: Magnification formula

Lateral magnification by objective:

M_o = \frac{v_o}{|u_o|} \approx \frac{f_o + L}{f_o} = 1 + \frac{L}{f_o}

Since $L \gg f_o$ in practice, $1$ is negligible:

\boxed{M_o \approx \frac{L}{f_o}}

Why this makes sense: A shorter $f_o$ means the objective is more "powerful" — it bends light more sharply, creating a larger image at the fixed tube length.

4. Derivation of Eyepiece Magnification $M_e$

Step 1: The eyepiece as a simple magnifier

The eyepiece takes the real image from the objective and acts like a magnifying glass. For relaxed eye (final image at infinity), the real image must be placed at the focal point of the eyepiece.

Step 2: Angular magnification

Angular magnification is defined as:

M_e = \frac{\text{angle subtended by image at eye}}{\text{angle subtended by object at near point}}

For a simple magnifier with image at infinity:

M_e = \frac{D}{f_e}

Where $D = 25$ cm (standard near point). …

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With the $f = 9\ \text{cm}$ magnifier of Exercise 9.22, maximum magnifying power occurs with the image at the near point ( $v=-25\ \text{cm}$ ): the lens is held $\tfrac{225}{34}\approx6.6\ \text{cm}$ from the card, the linear magnification is $\tfrac{34}{9}\approx3.8$ , and it equals the magnifying power in this case.

(a) Distance for maximum magnifying power

A simple magnifier gives its largest angular magnification when the virtual image forms at the near point, $D=25\ \text{cm}$ , because the object can then sit closest to the lens while staying in focus. Setting $v=-25\ \text{cm}$ (virtual, same side as object) with $f=+9\ \text{cm}$ :

\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-25} - \frac{1}{9} = -\frac{9 + 25}{225} = -\frac{34}{225} \Rightarrow u = -\frac{225}{34} \approx -6.6\ \text{cm}.

The lens should be held about $6.6\ \text{cm}$ from the card sheet.

(b) Magnification

m = \frac{v}{u} = \frac{-25}{-225/34} = \frac{25 \times 34}{225} = \frac{34}{9} \approx 3.8.

(c) Magnification versus magnifying power

The magnifying power with the image at the near point is

$M = 1 + \frac{D}{f} = 1 + \frac{25}{9} = \frac{34}{9} \approx 3.8.$ …

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Method: Angular Magnification (Magnifying Power) for a Simple Microscope (Single Convex Lens)

We use the lens formula and the definition of magnifying power for a simple microscope when the image is formed at the near point (D = 25 cm) — this gives the maximum possible magnifying power.

Step-by-step solution

Step 1: Identify given data (from Exercise 9.22 context)

Focal length of lens, $f = 5 \text{ cm}$ (assumed standard for such problems)
Least distance of distinct vision, $D = 25 \text{ cm}$
Object is a card sheet with squares of size $0.5 \text{ mm}$

Step 2: Condition for maximum magnifying power

The maximum magnifying power occurs when the final image is formed at the near point (D).

For a convex lens, the image is virtual and on the same side as the object.

Step 3: Apply lens formula

Lens formula:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

Here:

$v = -D = -25 \text{ cm}$ (negative because image is virtual)
$f = +5 \text{ cm}$ (convex lens)

Substitute:

\frac{1}{-25} - \frac{1}{u} = \frac{1}{5}

\Rightarrow -\frac{1}{25} - \frac{1}{u} = \frac{1}{5}

\Rightarrow -\frac{1}{u} = \frac{1}{5} + \frac{1}{25} = \frac{5 + 1}{25} = \frac{6}{25}

\Rightarrow \frac{1}{u} = -\frac{6}{25}

\Rightarrow u = -\frac{25}{6} \approx -4.17 \text{ cm}

Step 4: Answer (a)

The lens should be held at a distance of $4.17$ cm from the card sheet.

(The negative sign indicates the object is on the same side as the incoming light — standard sign convention.)

Step 5: Calculate magnification (b)

Linear magnification for a simple microscope:

m = \frac{v}{u} = \frac{-25}{-25/6} = 6

Magnification = 6 (the image is 6 times larger linearly). …

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Common Mistakes: Microscope Magnification (Exercise 9.22)

Mistake 1: Confusing Magnification with Magnifying Power

The error: Students treat magnification ( $m$ ) and magnifying power ( $M$ ) as the same thing in all contexts.

Why it happens: Both involve "how big" the image appears, but they measure different things:

Magnification ( $m$ ) = $\frac{\text{image height}}{\text{object height}} = \frac{v}{u}$ (purely geometric)
Magnifying power ( $M$ ) = $\frac{\text{angle subtended by image at eye}}{\text{angle subtended by object at near point}}$ (visual perception)

How to avoid: Always check what the question asks. In part (c), the answer is No — they are equal only when the image is formed at the near point (25 cm). For relaxed eye (image at infinity), $M \neq m$ .

Mistake 2: Using the Wrong Formula for Maximum Magnifying Power

The error: Students use $M = 1 + \frac{D}{f}$ when the question asks for "maximum possible magnifying power."

Why it happens: There are two standard formulas:

Image at near point (maximum $M$ ): $M_{\text{max}} = 1 + \frac{D}{f}$
Image at infinity (relaxed eye): $M_{\infty} = \frac{D}{f}$

How to avoid: "Maximum possible magnifying power" always means the image is at the near point ( $v = -25$ cm). Use $M = 1 + \frac{D}{f}$ .

Mistake 3: Forgetting Sign Conventions in Lens Formula

The error: Plugging $u$ and $v$ without signs into $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ .

Why it happens: The Cartesian sign convention is often ignored.

How to avoid: For a convex lens used as a simple microscope:

$u$ is negative (object on left)
$v$ is negative (virtual image on same side as object)
$f$ is positive (convex lens)

So for image at near point: $v = -25$ cm, solve $\frac{1}{f} = \frac{1}{-25} - \frac{1}{u}$ .

Mistake 4: Using $D = 25$ cm Without Checking Units

The error: Mixing cm and m in calculations.

Why it happens: $D$ (least distance of distinct vision) is usually 25 cm, but $f$ might be given in cm or m.

How to avoid: Convert everything to the same unit before calculating. If $f = 5$ cm, keep $D = 25$ cm. If $f = 0.05$ m, use $D = 0.25$ m.

--- …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL4 marks
$Q. A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.$
›Reveal solutionSolution

First find where the objective lens forms its (real, magnified) intermediate image using the lens formula; then require the eyepiece to form the final image at the near point (25 cm) for maximum magnification, which fixes the eyepiece's object distance; the lens separation is the sum of the two image/object distances, and total magnification is the product of the two individual magnifications.

Given: near point D = 25 cm, objective focal length f_o = 8.0 mm = 0.8 cm, object distance from objective u_o = 9.0 mm = 0.9 cm (object just outside focal length, as required for a microscope objective), eyepiece focal length f_e = 2.5 cm.

Step 1 - Image formed by objective: Using the thin lens formula (sign convention: distances measured from the lens, object distance taken negative):

u_o = -0.9 cm, f_o = +0.8 cm (convex lens)

1/v_o - 1/u_o = 1/f_o

1/v_o = 1/f_o + 1/u_o = 1/0.8 + 1/(-0.9) = 1.25 - 1.111 = 0.139

v_o = 1/0.139 = 7.2 cm

So the objective forms a real, inverted, magnified image at 7.2 cm on the far side of the objective.

Step 2 - Eyepiece positioned for image at near point (maximum magnification, image at D): This intermediate image (from the objective) acts as the object for the eyepiece. For maximum magnifying power, the eyepiece should form its final (virtual) image at the near point, i.e. v_e = -25 cm (virtual image, same side as the object, by sign convention).

1/v_e - 1/u_e = 1/f_e

1/(-25) - 1/u_e = 1/2.5

-0.04 - 1/u_e = 0.4

-1/u_e = 0.44

u_e = -1/0.44 = -2.27 cm

So the intermediate image must be 2.27 cm in front of the eyepiece for the eyepiece to form the final image at the near point.

…

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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL4 marks
$Q. Draw the ray diagram for the formation of image by a compound microscope and obtained the formula for magnification.$
›Reveal solutionSolution
Figure — The stem instructs 'Draw the ray diagram for the formation of image by a compound microscope'; the catalog fig

A compound microscope's objective forms a real, magnified image which then acts as the object for the eyepiece, which forms a further-magnified virtual image; total magnification is the product of the two individual magnifications.

Ray diagram (described): The object AB is placed just beyond the focus Fo of the objective lens (between fo and 2fo). The objective forms a real, inverted, magnified image A'B' beyond 2Fo on the other side. This intermediate image A'B' falls within the focal length of the eyepiece (between the eyepiece and its focus Fe). The eyepiece then acts as a simple magnifier on this intermediate image, producing a final virtual, further-magnified, and (relative to the object) inverted image A''B'' — formed at the near point D for maximum magnification, or at infinity for normal (relaxed-eye) adjustment. Both lenses are convex, mounted coaxially at opposite ends of a tube, with the objective having a small aperture and focal length, and the eyepiece having a larger aperture.

…

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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL4 marks
$Q. A compound microscope consists of an objective lens of focal length 2.0 cm and an eye piece of focal length 6.25 cm separated by a distance 15 cm. How far from the objective should an object be placed in order to obtain the final image at$
a) the least distance of distinct vision (25 cm) and
b) at infinity? What is the magnifying power of microscope in each case?
›Reveal solutionSolution
Work back from the eyepiece: (a) for image at 25 cm, u_e = -5 cm, objective forms image at 10 cm, u_o = -2.5 cm, M = 20; (b) for image at infinity, u_e = -6.25 cm, objective image at 8.75 cm, u_o = -2.59 cm, M = 13.5.

Data: f_o = 2.0 cm, f_e = 6.25 cm, tube separation L = 15 cm, D = 25 cm.

Final image at least distance of distinct vision (25 cm): Eyepiece: image virtual at v_e = -25 cm. 1/v_e - 1/u_e = 1/f_e: 1/(-25) - 1/u_e = 1/6.25 => -1/u_e = 1/6.25 + 1/25 = 5/25 = 1/5 => u_e = -5 cm. So the objective's image is 5 cm in front of the eyepiece; distance from objective = L - 5 = 10 cm => v_o = +10 cm. Objective: 1/v_o - 1/u_o = 1/f_o: 1/10 - 1/u_o = 1/2 => -1/u_o = 1/2 - 1/10 = 2/5 => u_o = -2.5 cm. Magnifying power: M = m_o (1 + D/f_e) = (v_o/|u_o|)(1 + D/f_e) = (10/2.5)(1 + 25/6.25) = 4 x 5 = 20.

Final image at infinity: …
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Q.(a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power?

(a) Distance for maximum magnifying power

(b) Magnification

(c) Magnification versus magnifying power

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